Lecture 12

Advanced Stages of Stellar Evolution – II Silicon Burning and NSE

The solution of reaction networks.

Consider the reaction pair:

dY1

dt=−Y

1λ

1+ Y

2λ

2

dY2

dt= Y

1λ

1− Y

2λ

2

An explicit solution of the linearized equations would be

Ynew

=Yold+δY where

δY1= (−Y

1λ

1+ Y

2λ

2) Δt

δY2= (Y

1λ

1− Y

2λ

2) Δt

In the usual case that the two species were not in equlibrium

Y1λ

1≠ Y

2λ

2, a large time step, Δt →∞, would lead to a

divergent value for the change in Y, including negative values.

For large t, the answer could oscillate.

On the other hand, forward or "implicit" differencing would giveδY1Δt

= − Y1 +δY1( )λ1 +   Y2 +δY2( )λ2δY2Δt

= − Y2 +δY2( )λ2 +   Y1 +δY1( )λ1

δY11Δt

+ λ1⎛⎝⎜

⎞⎠⎟ +δY2 −λ2( ) = −Y2 λ2 + Y1λ1

δY1 −λ1( ) +δY2 1Δt + λ2⎛⎝⎜

⎞⎠⎟ = Y2 λ2 − Y1λ1

Add equations ⇒δY1= −δY2; substituting, one also has :

δY1=Y1λ1 −Y2λ2

1 / Δt + λ1+ λ2

⎛⎝⎜

⎞⎠⎟

(if 1/Δt >> λ, same as the explicit solution)

Even if Δt →∞ the change in Y is finite and converges on the equilibrium value Y1λ1=Y2λ2

In general an n x n matrix n = 2 here

Silicon Burning

Silicon burning proceeds in a way different from any nuclear process discussed so far. It is analogous, in ways, to neon burning in that it proceeds by photodisintegration and rearrangement, but it involves many more nuclei and is quite complex. The reaction 28Si + 28Si ! (56Ni)* does not occur owing to the large Coulomb inhibition. Rather a portion of the silicon (and sulfur, argon, etc.)

melt by photodisintegration reactions into a sea of neutrons, protons, and alpha-particles. These lighter constituents add onto the remaining silicon and heavier elements, gradually increasing the mean atomic weight until species in the iron group are most abundant.

Initial Composition for Silicon Burning

The initial composition is mostly Si and S but which isotopes depends on whether one is discussing the inner core or locations farther out in the star. It is quite different for silicon core burning in a presupernova star and the explosive variety of silicon burning we will discuss later. In the center of the star, one typically has, after oxygen burning, and a phase of electron capture that goes on between oxygen depletion and silicon ignition: 30Si, 34S, 38Ar and a lot of other less abundant nuclei Farther outside of the core where silicon might burn explosively, one has species more characteristically with Z = N 28Si, 32S, 36Ar, 40Ca, etc. Historically, Si burning was discussed for a 28Si rich composition

Neutron excess after oxygen core depletion in 15 and 25 solar mass stars. The inner core is becoming increasingly neutronized , especially the lower mass stars.

15

25

Quasi-equilibrium This term is used to describe a situation where groups of adjacent isotopes, but not all isotopes globally, have come into equilibrium with respect to the exchange of n, p, , and .

It begins in neon burning with 20Ne + γ 16O + α andcontinues to characterize an increasing number of nuclei duringoxygen burning. In silicon burning, it becomes the rulerather than the exception. A typical "quasiequilibrium cluster" might include the equilibrated reactions:

28Si 29Si 30Si 31P 32S 28Si n n p p α

By which one means Y(28Si) Yn ρλnγ (28Si) ≈ Y(29Si) λγ n (

29Si)

Y(30Si)Ynρλnγ (29Si) ≈Y(30Si)λγ n (

30Si)

etc.

24 45 46 60 (at least)A A≤ ≤ ≤ ≤

Late during oxygen burning, many isolated clusters grow and merge until, at silicon ignition, there exist only two large QE groups

Reactions below 24Mg, e.g., 20Ne(,)24Mg and 12C(,)16O are, in general, not in equilibrium with their inverses (exception, 16O(,)20Ne which has been in equilibrium since neon burning). Within the groups heavier than A = 24, except at the boundaries of the clusters, the abundance of any species is related to that of another by successive application of the Saha equation.

40 32 36 40

28 28 32 36

28 32 36

32 36 40

3 3

( ) ( ) ( ( ). .,

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( , ) .

Y Ca Y S Y Ar Y Cae g

Y Si Y Si Y S Y Ar

Y Si Y S Y Ar

S Ar Ca

f T Q Y etc

α αγ α αγ α αγ

γα γα γα

αγ α

ρ λ ρ λ ρ λλ λ λ

ρ

⎛ ⎞⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞⎛ ⎞

= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

=

The situation at the end of oxygen burning is that there are two large QE groups coupled by non-equilibrated links near A = 45.

Early during silicon burning these two groups merge and the only remaining non-equilibrated reactions are for A < 24.

45A >

24 45A≤ ≤

The non-equilibrated link has to do with the double shell closure at Z = N = 20

Y ( A Z ) = C( A Z ,ρ,T9 )Y (28 Si)Yα

δαYpδ pYn

δ n

where δα = largest integer ≤Z-14

2δ n = N −14− 2δαδ p = Z −14− 2δα e.g.,

35 Cl 17 protons; 20 neutrons

δα =1 δ p = 1 δ n = 240K 19 protons 21 neutrons

δα = 2 δ p = 1 δ n = 3

Within that one group, which contains 28Si and the vast majority of the mass, one can evaluate any abundance relative to e.g., 28Si

Need 6 parameters: Y, Yp, Yn and Y(28Si) plus T and , but …

C(A Z ) = ρNA( )δα +δ p +δn

C '(A Z )

C '(A Z ) = 5.942×1033T93/2( )

− δα +δ p +δn( )

G(A Z )

G(28Si)2− δ p +δn( ) A

28

⎛⎝⎜

⎞⎠⎟

3/21

4

⎛⎝⎜

⎞⎠⎟

3δα /2

exp(Q / kT )

where

Q = BE(A Z ) − BE(28Si) − δαBE(α )

i.e., the energy required to dissociate the nucleus AZ into 28Si and alpha particles. The binding energy of a neutron or proton is zero.

32S 31P 28Si 29Si 30Si

32 30 2

30 28 2

32 28

32 32 302 2

28 30 28

For constant and T

( ) ( )

( ) ' ( )

( ) '' ( )

( ) ( ) ( )

( ) ( ) ( )

p

n

p n

Y S const Y Si Y

Y Si const Y Si Y

Y S const Y Si Y

Y S Y S Y SiY Y Y

Y Si Y Si Y Si

α

α

ρ

=

=

=

⎛ ⎞ ⎛ ⎞⎛ ⎞∝ ∝ ∝⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

i

i

i

Yα= C

α(T

9,ρ) Y

n

2Y

p

2

Cα=

1

2(5.94 ×10

33)−3ρ

3T

9

9/2exp(328.36 / T

9)

where 328.36 = BE(α )/kT

This reduces the number of independent variables to 5, but wait …

Moreover there exist loops like:

p

p

n n

% QE

24 A 60≤ ≤

28Si

24Mg+%

20Ne+%

16O+%

12C+%

3%

7 !

28

9, , , ( ) QET Y Siρ η ⇒

The cluster evolves at a rate given by 24Mg(,)20Ne

The photodisintegration of 24Mg provides a s (and n s and p s since Ya =CaYn2Yp2) which add onto the QE group gradually increasing its mean atomic weight. As a result the intermediate mass group, Si-Ca gradually melts into the iron group.

Y ( A Z ) = C( A Z ,ρ,T9 )Y (

28 Si)YαδαYp

δ pYnδ n

60 24

60 24

1

( )

i i

A

i i i

A

AY

N Z Y η

≥ ≥

≥ ≥

≈

− ≈

∑

∑

The large QE cluster that includes nuclei from A = 24 through at least A = 60 contains most of the matter (20Ne, 16O, 12C, and !are all small), so we have the additional two constraints

The first equation can be used to eliminate one more unknown, say Yp, and the second can be used to replace Yn with an easier to use variable, . Thus 4 variables now specify the abundances of all nuclei heavier than magnesium . These are , T9, , and Y(28Si)

mass conservation charge conservation

Nature of the burning: Lighter species melt away while the iron group grows

But this assumes 28Si burns to 56Ni ( = 0.002 approximation)

This is all a bit misleading because, except explosively (later), silicon burning does not produce 56Ni. There has been a lot of electron capture during oxygen burning and more happens in silicon burning.

E.g., Si ignition in a 15 M

star ηc≈0.07 Y

e≈ 0.46

Si depletion ηc ≈0.13 Ye ≈ 0.44

Under these conditions silicon burning produces 54Fe, 56Fe, 58Fe and

other neutron rich species in the iron group.

Suppose 56

30

⎛⎝⎜

⎞⎠⎟

30Si → 56Fe

qnuc = 9.65 × 1017 492.26( ) / 56− 255.62( ) / 30⎡⎣ ⎤⎦

= 2.6 × 1017 erg g-1 which is closer to correct for Si

core burning than 1.9 × 1017 erg g−1

i.e., qnuc = 9.65×1017 Xi

AiBE(Ai )∑

This is very like neon burning except that 7 alpha-particles are involved instead of one.

Reaction rates governing the rate at which silicon burns:

Generally speaking, the most critical reactions will be those connecting equilibrated nuclei with A > 24 (magnesium) with alpha-particles. The answer depends on temperature and neutron excess: Most frequently, for small, the critical slow link is 24Mg(,)20Ne The reaction 20Ne(,)16O has been in equilibrium with 16O(,)20Ne ever since neon burning. At high temperatures and low Si-mass fractions, 20Ne(,)24Mg equilibrates with 24Mg(,)20Ne and 16O(,)12C becomes the critical link. However for the values of actually appropriate to silicon burning in a massive stellar core, the critical rate is 26Mg(p,)23Na(p,)20Ne

To get 24Mg

To get %

Nucleosynthesis

Basically, silicon burning in the star s core turns the products of oxygen burning (Si, S, Ar, Ca, etc.) into the most tightly bound nuclei (in the iron group) for a given neutron excess, .!!   The silicon-burning nucleosynthesis that is ejected by a super- nova is produced explosively, and has a different composition dominated by 56Ni. The products of silicon-core and shell burning in the core are both so neutron- rich ( so large) that they need to be left behind in a neutron star or black hole. However, even in that case, the composition and its evolution is critical to setting the stage for core collapse and the supernova explosion that follows.

Following Si-burning at the middle of a 25 solar mass star:

54Fe 0.487 58Ni 0.147 56Fe 0.141 55Fe 0.071 57Co 0.044

Neutron-rich nuclei in the iron peak. Ye = 0.4775

Following explosive Si-burning in a 25 solar mass supernova, interesting species produced at Ye = 0.498 to 0.499.

44Ca 44Ti 47,48,49Ti 48,49Cr 51V 51Cr 55Mn 55Co 50,52,53Cr 52,53Fe 54,56,57Fe 56,57Ni 59Co 59Cu 58,60,61,62Ni 60,61,62Zn

product parent

Silicon burning nucleosynthesis

44Ti and 56.57Ni are important targets of -ray astronomy

25 M

η = neutron excess

25 Solar Mass Pop I

PreSupernova

(Woosley and Heger 2007)

Explosive burning will happen here

This part will collapse and make a shock

Nuclear Statistical Equilibrium

24 24 24 20

16 20 20 16

12 16 16 12

12 12

( , ) ( , )

( , ) ( , ) (for a long time already)

( , ) ( , )

3 ( , )2

Ne Mg Mg Ne

O Ne Ne O

C O O C

C C

α γ γ α

α γ γ α

α γ γ α

α γ α α

↔

↔

↔

→ ↔

As the silicon abundance tends towards zero (though it never becomes microscopically small), the unequilibrated reactions below A = 24 finally come into equilibrium

The 3 reaction is the last to equilibrate. Once this occurs, every isotope is in equilibrium with every other isotope by strong and electromagnetic reactions (but not by weak interactions)

In particular, Y(28Si) = f (T ,ρ)Yα7 with the result that now

only 3 variables, ρ, T9,and η now specify the abundances

of everything

Y(AZ ) = C(A Z ,ρ,T9) Y

n

NYp

Z

C(A Z ,ρ,T9) = ρN

A( )A−1

C '(A Z ,T9)

C ' (A Z ,T9)=

G(A Z ,T9)

2Aθ1−A exp BE(A Z ) / kT⎡⎣ ⎤⎦

θ =5.943 × 1033 T9

3/2

G(A Z ,T9) is the temperature-dependent

partition function.

At low T

G(AZ ,T9) = 2J

i+1( )∑ e−Ei /kT

At high T, though (see earlier discussion of nuclear level density)

G(AZ ,T9)≈

π6akT

ea(kT )

a ≈A

9MeV-1

Until the temperature becomes very high (kT >> 1 MeV)The most abundant nuclei are those with large binding energyper nucleon and "natural" values of η. For example,

η = 0 56Ni 0.037 54Fe

0.071 56Fe etc.

In general, the abundance of an isotope peaks at itsnatural value for η. E. g.,

η(56Fe) = N − ZA

= 30 - 2656

=0.0714The resultant nucleosynthesis is most sensitive to .!

True Equilibrium

If the weak interactions were also to be balanced, (e.g., neutrino capture occurring as frequently on the daughter nucleus as electron capture on the parent), one would have a state of true equilibrium. Only two parameters, and T, would specify the abundances of everything. The last time this occurred in the universe was for temperatures above 10 billion K in the Big Bang. However, one can also have a dynamic weak equilibrium where neutrino emission balances anti-neutrino emission, i.e., when

0e

dY

dt=

This could occur, and for some stars does, when electron-capture balances beta-decay globally, but not on individual nuclei. The abundances would be set by and T, but would also depend on the weak interaction rate set employed.

Weak Interactions

Electron capture, and at late times beta-decay, occur for a variety of isotopes whose identity depends on the star, the weak reaction rates employed, and the stage of evolution examined. During the late stages it is most sensitive to eta, the neutron excess. Aside from their nucleosynthetic implications, the weak interactions determine Ye, which in turn affects the structure of the star. The most important isotopes are not generally the most abundant, but those that have some combination of significant abundance and favorable nuclear structure (especially Q-value) for weak decay. From silicon burning onwards these weak decays provide neutrino emission that competes with and ultimately dominates that from thermal processes (i.e., pair annihilation).

He – depletion O – depletion PreSN

25 M

solar metallicity

He – depletion O – depletion PreSN

25 M

zero initial metalicity

The distribution of neutron excess, , within two stars of 25 solar masses (8 solar mass helium cores) is remarkably different. In the Pop I star, is approximately 1.5 x 10-3 everywhere except in the inner core (destined to become a collapsed remnant)

In the Pop III (Z = 0) star the neutron excess is essentially zero at the end of helium burning (some primordial nitrogen was created) Outside of the core is a few x 10-4, chiefly from weak interactions during carbon burning. Note some primary nitrogen production at the outer edge where convection has mixed 12C and protons.

log %

Interior mass

Interior mass

log %

Si-depletion Si-shell burn Core contraction PreSN T(109 K) 3.78 4.13 3.55 7.16 (g cm-3) 5.9 x 107 3.2 x 108 5.4 x 108 9.1 x 109 Ye 0.467 0.449 0.445 0.432 e-capture 54,55Fe 57Fe, 61Ni 57Fe, 55Mn 65Ni, 59Fe -decay 54Mn, 53Cr 56Mn,52V 62Co,58Mn 64Co, 58Mn

O-depletion O-shell Si-ignition Si-shell T(109K) 2.26 1.90 2.86 3.39 (g cm-3) 1.2 x 107 2.8 x 107 1.1 x 108 4.5 x 107

Ye 0.498 0.495 0.489 0.480 e-capture 35Cl, 37Ar 35Cl, 33S 33S,35Cl 54,55Fe -decay 32P, 36Cl 32P,36Cl 32P, 28Al 54,55Mn

15 solar mass star (Heger et al 2001)

Fe

Si, S, Ar, Ca

He, C

H, He He

1400 R

0.5 R

0.1 R

0.003 R

25 M

Presupernova Star (typical for 9 - 130 M

)

O,Mg,Ne

C,O

240,000 L

(note scale breaks)

As silicon shells, typically one or at most two, burn out, the iron core grows in discontinuous spurts. It approaches instability. Pressure is predominantly due to relativistic electrons. As they become increasingly relativistic, the structural adiabatic index of the iron core hovers precariously near 4/3. The presence of non- degenerate ions has a stabilizing influence, but the core is rapidly losing entropy to neutrinos making the concept of a Chandrasekhar Mass relevant. In addition to neutrino losses there are also two other important instabilities: •  Electron capture – since pressure is dominantly from electrons,

removing them reduces the pressure. •  Photodisintegration – which takes energy that might have provided pressure and uses it instead to pay a debt of negative nuclear energy generation.

See also Clayton Fig 7-9 and discussion

Illiadis

Typical presupernova central conditions

The transition from the iron group to He and especially from He to 2n + 2p absorbs an enormous amount of energy. Figure needs correcting for high temperature nuclear partition functions.

Dominant constituent in NSE ( approximately 0)

Photodisintegration:

Yα = Cα (T9 ,ρ) Yn2Yp

2

Cα=

1

2(5.94 × 1033 )−3 ρ 3T

9

9/2 exp(328.36 / T9)

where 328.36 = BE(α )/kT ⇒ Cα increases rapidly as T ↓

from page 12 of this lecture

Setting Y = 1/8 (i.e., X = ½ divided by 4) and Yp=Yn=1/4 gives the line for the helium-nucleon transition on the previous page. The Ni- transition is given by Clayton problem 7-11 and eqn 7-22. It comes from solving the Saha equation for NSE (page 30 of these notes) for the case

X(56Ni)= Xα =0.5⇒Y (

56Ni)=1/112; Yα =1/ 8

Y (56Ni)=C56(T9 )ρ

12Yα13

The photodisintegration of one gram of 56Ni to one gram of -particles absorbs:

qnuc

= 1.602×10−6 NA

(δYi)(BE

i) − qν erg/gm∑

= 9.64×1017 −1

56

⎛⎝⎜

⎞⎠⎟

483.993( )+1

4

⎛⎝⎜

⎞⎠⎟

28.296( )⎡

⎣⎢

⎤

⎦⎥

= −1.51×1018 erg g−1

Similarly, the photodisintegration of one gram of ’s to one gram of nucleons absorbs:

qnuc

=9.64×1017 −1

4

⎛⎝⎜

⎞⎠⎟

28.296( ) + 0⎡

⎣⎢⎤

⎦⎥

=−6.82×1018 erg g−1

essentially undoing all the energy released by nuclear reactions since the zero age main sequence.

Ent

ropy

(S/N

Ak)

What about degeneracy? Can the core be supported by electron degeneracy pressure and form a stable (Fe) dwarf?

Ent

ropy

Because of increasing degeneracy the concept of a Chandrasekhar Mass for the iron core is relevant – but it must be generalized.

0 implies degeneracyη >

The Chandrasekhar Mass

2

3 3 4 4 3 2

4/3

4 4

15 -2

4/3

Traditionally, for a fully relativistic, completely degenerate gas:

20.745

1.2435 10 dyne cm

1.457 M at 0.5

0

5

83

.

c c e

Ch

c c

e

eM Y

P K Y G M

K

Y

ρ

ρ ρ= =

= ×

= =

=

This is often referred to loosely as 1.4 or even 1.44 solar masses

see lecture 1

BUT 1)  Ye here is not 0.50 (Ye is actually a function

of radius or interior mass)

2)  The electrons are not fully relativistic in the outer layers ( is not 4/3 everywhere)

3)  General relativity implies that gravity is stronger than classical and an infinite central density is not allowed (there exists a critical for stability)

4)  The gas is not ideal. Coulomb interactions reduce the pressure at high density

5)  Finite temperature (entropy) corrections

6)  Surface boundry pressure (if WD is inside a massive star)

7)  Rotation

Effect on MCh Relativistic corrections, both special and general, are treated by Shapiro and Teukolsky in Black Holes, White Dwarfs, and Neutron Stars pages 156ff. They find a critical density (entropy = 0).

2

10 -30.502.646 10 gm cm

c

eY

ρ⎛ ⎞

= × ⎜ ⎟⎝ ⎠

Above this density the white dwarf is (general relativistically) unstable to collapse. For Ye = 0.50 this corresponds to a mass

2/31

1.415 M

in general, the relativistic correction to the Newtonian value is

M 0.50 9 4 310

M

2.87% 0.50

2.67% 0.45

Ch

e

e

e

M

Y

Y

Y

−

=

⎡ ⎤⎛ ⎞Δ ⎢ ⎥= − + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦=− =

=− =

Coulomb Corrections

Three effects must be summed – electron-electron repulsion, ion-ion repulsion and electron ion attraction. Clayton p. 139 – 153 gives a simplified treatment and finds, over all, a decrement to the pressure (eq. 2-275)

ΔPCoul

= −3

10

4π3

⎛⎝⎜

⎞⎠⎟

Z2/3

e2

ne

4/3

Fortunately, the dependence of this correction on ne is the same as relativistic degeneracy pressure. One can then just proceed to use a corrected

1/32 5/3

0 2/3

4 /3 4 /3

0 3 2/3

4 /3

2 31

5

1 4.56 10

eK K Z

c

K Z

π−

⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤= − ×⎣ ⎦

( )1/3

0 2 4/3

4 /3

3/ 2

Ch

0 0

Ch

2/3

0

Ch

where 34

Mand

M

hence

M 1 0.02266

A

Ch

cK N

K

K

ZM

π=

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

2

e

12

Ch e

56

Putting the relativistic and Coulomb corrections together

with the dependence on Y one has

M 1.38 M for C (Y 0.50)

= 1.15 M for Fe (Y

= =

e

e

260.464)

56

= 1.08 M for Fe-core with ≈

So why are iron cores so big at collapse (1.3 - 2.0 M

) and

why do neutron stars have masses ≈1.4 M

?

( )

( )

2 2 24/30

4/3 2 2 4

1/31/3

For 3, 4 / 3, relativistic degeneracy

1 21 1 ...

30.01009

8

(Clayton 2-48)

c e

e

Fe e

e

n

k TP K Y

x m c

phx n Ymc m c

γ

πρ

ρπ

= =

⎡ ⎤⎛ ⎞= + − +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

Finite Entropy Corrections

Chandrasekhar (1938) Fowler & Hoyle (1960) p 573, eq. (17) Baron & Cooperstein, ApJ, 353, 597, (1990)

In particular, Baron & Cooperstein (1990) show that

2

1/33

1/3

7

3/ 2

4 /3

2

0

21 ...

3

3

8

1.11( ) MeV

and since a first order expansion gives

1

o

F

eF F

F e

Ch

Ch Ch

F

kTP P

h np c

Y

M K

kTM M

πε

επ

ε ρ

πε

⎛ ⎞⎛ ⎞⎜ ⎟= + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞= =⎜ ⎟

⎝ ⎠=

∝

⎛ ⎞⎛ ⎞⎜ ⎟≈ + ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

And since, in the appendix to these notes we show

se=π 2kTY

e

εF

(relativistic degeneracy)

one also has

MCh≈ M

Ch

0 1+s

e

πYe

⎛

⎝⎜⎞

⎠⎟

2

+ ...⎛

⎝⎜⎜

⎞

⎠⎟⎟

The entropy of the radiation and ions also affects MCh, but much less. This finite entropy correction is not important for isolated white dwarfs. They re too cold. But it is very important for understanding the final evolution of massive stars.

Because of its finite entropy (i.e., because it is hot) the iron core develops a mass that, if it were cold, could not be supported by degeneracy pressure. Because the core has no choice but to decrease its electronic entropy (by neutrino radiation, electron capture, and photodisintegration), and because its (hot) mass exceeds the Chandrasekhar mass, it must eventually collapse.

E.g., on the following pages are excerpts from the final day in the

life of a 15 M

star. During silicon shell burning, the electronic entropy

ranges from 0.5 to 0.9 in the Fe core and is about 1.3 in the convective

shell.

MCh ≈1.08 M 1+0.7

π (0.45)⎛⎝⎜

⎞⎠⎟

2⎛

⎝⎜

⎞

⎠⎟ = 1.34 M > 0.95 M

The Fe core plus Si shell is also stable because

MCh≈ 1.08 M 1+1.0

π (0.47)⎛⎝⎜

⎞⎠⎟

2⎛

⎝⎜

⎞

⎠⎟ = 1.57 M > 1.3M

But when Si burning in this shell is complete:

Neutrino losses farther reduce se. So too do photodisintegration and electron capture as we shall see. And the boundry pressure of the overlying silicon shell is not entirely negligible. The core collapses

e

e

Ch

3) The Fe core is now ~1.3 M .

s central = 0.4

s at edge of Fe core = 1.1

average 0.7

M now about 1.34 M (uncertain to at least a

≈

few times 0.01 M

Electron

The collapse begins on a Kelvin-Helmholz (neutrino) time scale and accelerates to a dynamic implosion as other instabilities are encountered. Photodisintegration: As the temperate and density rise, the star seeks a new fuel to burn, but instead encounters a phase transition in which the NSE distribution favors particles over bound nuclei. In fact, this transition never goes to completion owing to the large statistical weight afforded the excited states of the nuclei. But considerable energy is lost in a partial transformation. Recall for s.

56

17

56

18 -1

13 "4 "

28.296 492.2629.65 10

4 56

1.7 10 erg g

nuc

photo

Fe n

q X

q

α→ +

⎛ ⎞≈ × − Δ⎜ ⎟⎝ ⎠

= − ×

not really free neutrons. They stay locked inside bound nuclei that are progressively more neutron rich.

i.e., BE/A for a typical iron group nucleus

What happens? As the density rises, so does the pressure (it never decreases at the middle), but so does gravity. The rise in pressure is not enough to maintain hydrostatic equilibrium, i.e., remains slightly less than 4/3. The collapse accelerates. Photodisintegration decreases se because at constant total entropy (the collapse is almost adiabatic), si increases since 14 particles have more statistical weight than one nucleus. The increase in si comes at the expense of se.

Electron capture The pressure and entropy come mainly from electrons, but as the density increases, so does the Fermi energy, F. The rise in F means more electrons have enough energy to capture on nuclei turning protons to neutrons inside them. This reduces Ye which in turn makes the pressure and entropy at a given density smaller.

( )1/3

71.11 MeV

F eYε ρ=

By 2 x 1010 g cm-3, F= 10 MeV which is above the capture threshold for all but the most neutron-rich nuclei. There is also briefly a small abundance of free protons (up to 10-3 by mass) which captures electrons.

But the star does not a) photodisintegrate to neutrons and protons; then b) capture electrons on free protons; and c) collapse to nuclear density as a free neutron gas as some texts naively describe. Bound nuclei persist, then finally touch and melt into one gigantic nucleus with 1057 nucleons – the neutron star. Ye declines to about 0.37 before the core becomes opaque to neutrinos. (Ye for an old cold neutron star is about 0.05; Ye for the neutron star that bounces when a supernova occurs is about 0.29). The effects of a) exceeding the Chandrasekhar mass, b) photodisintegration and c) electron capture operate together, not independently.

Appendix on Entropy

T

S = k log W

Zentralfriedhof Vienna, Austria

41

3

AN kT

P aTρ

µ= +

3/ 2 3 3/ 2( / ) /( / )T T Tρ ρ =

for ideal gas plus radiation

dividing by k makes s dimensionless

As discussed previously

34 4

2

34 4

2 3

3

3

rad

Eg., just the radiation part

1/

4 1 1

3

4 1

3

44

3

4

3

4So S

3

V

T dS d PdV

aTT dS dT aT d aT dV

aTdT aT dV aT dV

dS aT V dT aT dV

d aT V

aT

ρε

ρρ ρ

ρ

ρ

≡= +

⎛ ⎞= + − +⎜ ⎟

⎝ ⎠

= + +

= +

⎛ ⎞= ⎜ ⎟

⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

Cox and Guili (24.76b)

expression Cox and Guili Principles of Stellar Structure Second edition A. Weiss et al Cambridge Scientific Publishers

Reif Fundamentals of Statistical and Thermal Physics McGraw Hill

η=µ

kT

where µ, the chemical potential is defined by

T S =E + PV −µ N (CG10.20)

hence

Se

/V = (εeρ +P

e− µ n

e) / T

Se= ε

e+

Pe

ρ−µn

e

ρ⎛

⎝⎜⎞

⎠⎟/ T

Ye=

ne

ρ NA

se=S

e/ N

Ak =

εeρ +P

e

ρNAkT

⎛

⎝⎜⎞

⎠⎟−ηY

e

=5

2−η

⎛⎝⎜

⎞⎠⎟

Ye

η0

For an ideal gas

ε ρ + P=3

2+1

⎛⎝⎜

⎞⎠⎟ρN

AkT

For a non-relativistic, non-degenerate electron gas,

Clayton 2-63 and 2-57 imply (for η>1 (great degeneracy)

ne≈

8πc

3h

3kT( )

3 1

3η3 1+

π 2

η2⎡

⎣⎢

⎤

⎦⎥

Pe≈

8π3c3h3

kT( )4 1

4η4 1+

2π 2

η2+

7π 4

15η4⎡

⎣⎢

⎤

⎦⎥ Pe ∝ ne

4/3( )

Pe

nekT

≈1

4η

1+2π 2

η2

1+π 2

η2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

>>1

η π and keeping only terms in η−2

Pe

nekT

≈η4

1+2π 2

η2⎛

⎝⎜⎞

⎠⎟1−

π 2

η2⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

≈η4

1+2π 2

η2−π 2

η2⎛

⎝⎜⎞

⎠⎟=η4+π 2

4η

ue=3P

eη =

µ

kTµ ≈ε

F

Se

V= (u

e+ P

e− µ n

e) / T ≈

4Pe−µn

e( )T

≈n

ekT η +

π 2

η⎛

⎝⎜⎞

⎠⎟−ηn

ekT

T=π 2n

ek

η

=π 2ρN

AY

ek

η

se=

Se

ρNAk≈π 2Y

e

η

se≈π 2kTY

e

εF

and negligible radiation pressure (entropy)